College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter P - Prerequisites: Fundamental Concepts of Algebra - Exercise Set P.5 - Page 75: 125

Answer

We factor out the given polynomial by trial and error. Example: 3$x^{2}$ + 10 x + 8 First multiply coefficient of $x^{2}$ and constant term, then break the product of both into two factors so that the sum of both factor will be same as the product. Here 3$\times$8 = 24 factor of 24 = 4 and 6 that sum will be 10. 3$x^{2}$ + 10 x + 8 = 3$x^{2}$ + 6 x + 4 x + 8 Take 3$x$ and 4 common 3$x^{2}$ + 6 x + 4 x + 8 = 3 x(x + 2) + 4( x + 2) = (3 x + 4) (x + 2) So the factor of polynomial = (3 x + 4) (x + 2)

Work Step by Step

There are many methods to factor out a polynomial. We find out the factors of a given polynomial by trial and error. Example: 3$x^{2}$ + 10 x + 8 First multiply coefficient of $x^{2}$ and constant term, then break the product of both into two factors so that the sum of both factor will be same as the product. Here 3$\times$8 = 24 factor of 24 = 4 and 6 that sum will be 10. 3$x^{2}$ + 10 x + 8 = 3$x^{2}$ + 6 x + 4 x + 8 Take 3$x$ and 4 common 3$x^{2}$ + 6 x + 4 x + 8 = 3 x(x + 2) + 4( x + 2) = (3 x + 4) (x + 2) So the factor of polynomial = (3 x + 4) (x + 2)
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